Question

In the reaction \[{\text{A}} \to {\text{ product }}\], \[\dfrac{{{\text{ - dA}}}}{{{\text{dt}}}}{\text{ = K[A]}}\]. If we start with 10M of A, then after one natural lifetime, the concentration of A decreased to:

Answer 1

Hint: Reactions whose rate is determined by the change of one concentration term only is known as the first-order reaction. Use the first-order kinetic equation and determine the decrease in the concentration of reactant ‘A’ after one natural lifetime.

Complete step by step answer:
The reaction given to us is:
\[{\text{A}} \to {\text{ product }}\]
The rate reaction is \[\dfrac{{{\text{ - dA}}}}{{{\text{dt}}}}{\text{ = K[A]}}\]
As the rate of reaction depends on the change in concentration of the only reactant A so this is a first-order reaction.
The first-order kinetic equation is:
\[{\text{C = }}{{\text{C}}_{\text{0}}}{{\text{e}}^{{\text{ - Kt}}}}\]
Where,
\[{{\text{C}}_{\text{0}}}\]= initial concentration
\[{\text{C}}\]= concentration at time t
t = time
K = rate constant
After one natural lifetime that is at time t
\[{\text{t = }}\dfrac{{\text{1}}}{{\text{K}}}\]
Now, we can substitute 10M for initial concentration \[{{\text{C}}_{\text{0}}}\] and \[\dfrac{{\text{1}}}{{\text{K}}}\] for time t.
\[{\text{C = 10M }}{{\text{e}}^{{\text{ - K $\times$ }}\dfrac{{\text{1}}}{{\text{K}}}}}\]
\[{\text{C = }}\dfrac{{{\text{ 10}}}}{{\text{e}}}{\text{M }}\]
Thus, after one natural lifetime, the concentration of A decreased to \[\dfrac{{{\text{ 10}}}}{{\text{e}}}{\text{M }}\].

Hence, the correct option is (C) \[\dfrac{{{\text{ 10}}}}{{\text{e}}}{\text{M }}\].

Note: The rate of reaction is the change in concentration of reactant or product in unit time. In the case of first-order reaction rate of reaction depends on the change in concentration of only one reactant. The rate constant value for the given reaction will be always constant. It is independent of the change in concentration of reaction components.